The equations of the common tangents to the t | vedclass

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(B) Any tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.For this line to be a tangent to

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$y = \pm x \pm \sqrt{a^2 - b^2}$

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(B) Any tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.For this line to be a tangent to the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,we substitute $y = mx + c$ into the second equation:$\frac{(mx + c)^2}{a^2} - \frac{x^2}{b^2} = 1$$b^2(m^2x^2 + 2mcx + c^2) - a^2x^2 = a^2b^2$$x^2(b^2m^2 - a^2) + 2b^2mcx + (b^2c^2 - a^2b^2) = 0$.Since the line is a tangent,the discriminant of this quadratic equation must be zero:$D = (2b^2mc)^2 - 4(b^2m^2 - a^2)(b^2c^2 - a^2b^2) = 0$$4b^4m^2c^2 - 4b^2(b^2m^2 - a^2)(c^2 - a^2) = 0$$b^2m^2c^2 - (b^2m^2c^2 - a^2b^2m^2 - a^2c^2 + a^4) = 0$$a^2b^2m^2 + a^2c^2 - a^4 = 0$$b^2m^2 + c^2 - a^2 = 0 \Rightarrow c^2 = a^2 - b^2m^2$.Equating the two expressions for $c^2$:$a^2m^2 - b^2 = a^2 - b^2m^2$$m^2(a^2 + b^2) = a^2 + b^2$ $\Rightarrow m^2 = 1$ $\Rightarrow m = \pm 1$.Substituting $m^2 = 1$ into $c^2 = a^2m^2 - b^2$:$c^2 = a^2 - b^2 \Rightarrow c = \pm \sqrt{a^2 - b^2}$.Thus,the common tangents are $y = \pm x \pm \sqrt{a^2 - b^2}$.

The equations of the common tangents to the two hyperbolas $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ are-

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